Write the sum $LaTeX: \displaystyle 45+49+53 \ldots +141+145$ in sigma notation and then find the sum.

The common difference is given by $LaTeX: \displaystyle a_2-a_1=49-(45)=4$ . Using the first term gives the sequene $LaTeX: \displaystyle a_n= 45+(n-1)(4)$ . Setting the general term equal to the last term and solving for $LaTeX: \displaystyle n$ gives $LaTeX: \displaystyle 45+(n-1)(4)=145 \implies n = 26$ . Writing in sigma notation gives $LaTeX: \displaystyle \displaystyle \sum_{n=1}^{26} \left(4 n + 41\right)$ . Using the formula for a finite arithmetic sum gives $LaTeX: \displaystyle \frac{ 26(45+145) }{2}=2470$ .